ECE 576 – Power System Dynamics and Stability Lecture 18: Multimachine Simulation: Explicit Methods Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign overbye@illinois.edu

Announcements Read Chapter 7 Homework 5 is due today Homework 6 is due on April 10 A useful reference is B. Stott, "Power System Dynamic Response Calculations," Proc. IEEE, vol. 67, pp. 219-241 Another key reference is J.M. Undrill, "Structure in the Computation of Power-System Nonlinear Dynamic Response," IEEE Trans. Power App. and Syst., vol. 88, pp. 1-6, January 1969.

Question about HYGOV

Power Grid Physical Vulnerabilities About one year ago (4/16/13) there was a sniper attack on the PGE Metcalf substation Considered a sophisticated attack partially because they disabled communication first; shooting lasted for 19 minutes, with 17 transformers knocked out (some 500/230 kV ones) but they were not permanently damaged On 3/12/14 the Wall Street Journal (WSJ) reported that a power flow study by FERC reported that all three grids could be collapsed by knocking out 4 substations in the EI, 3 in WECC and 2 in ERCOT The WSJ was criticized by FERC, NERC and EEI for mentioning this vulnerability (though they did not mention the substations)

Euler's Solution At t = Tclear the fault is self-cleared, with the equations changing to The integration continues using the new equations

Euler's Solution Results (Dt=0.02) The below table gives the results using Dt = 0.02 for the beginning time steps This is saved as PowerWorld case B2_CLS_Infinite. The integration method is set to Euler's on the Transient Stability, Options, Power System Model page Time Gen 1 Rotor Angle, Degrees Gen 1 Speed (Hz) 23.9462 60 0.02 60.2 0.04 25.3862 60.4 0.06 28.2662 60.6 0.08 32.5862 60.8 0.1 38.3462 61 0.12 45.5462 60.8943 0.14 51.9851 60.7425 0.16 57.3314 60.5543 0.18 61.3226 60.3395 0.2 63.7672 60.1072 0.22 64.5391 59.8652 0.24 63.5686 59.6203 0.26 60.8348 59.3791 0.28 56.3641 59.1488

Generator 1 Delta: Euler's The below graph shows the generator angle for varying values of Dt; numerical instability is clearly seen

Second Order Runge-Kutta Runge-Kutta methods improve on Euler's method by evaluating f(x) at selected points over the time step One approach is a second order method (RK2) in which That is, k1 is what we get from Euler's; k2 improves on this by reevaluating at the estimated end of the time step Error varies with the cubic of the time step This is also known as Heun's method or as the Improved Euler's or Modified Euler's Method

Second Order Runge-Kutta (RK2) Again Assuming a time step Dt = 0.02 seconds, and a Tclear of 0.1 seconds, then using Heun's approach

RK2 Solution Results (Dt=0.02) The below table gives the results using Dt = 0.02 for the beginning time steps This is saved as PowerWorld case B2_CLS_Infinite. The integration method should be changed to Second Order Runge-Kutta on the Transient Stability, Options, Power System Model page Time Gen 1 Rotor Angle, Degrees Gen 1 Speed (Hz) 23.9462 60 0.02 24.6662 60.2 0.04 26.8262 60.4 0.06 30.4262 60.6 0.08 35.4662 60.8 0.1 41.9462 61 0.12 48.6805 60.849 0.14 54.1807 60.6626 0.16 58.233 60.4517 0.18 60.6974 60.2258 0.2 61.4961 59.9927 0.22 60.605 59.7598 0.24 58.0502 59.5343 0.26 53.9116 59.3241 0.28 48.3318 59.139

Generator 1 Delta: RK2 The below graph shows the generator angle for varying values of Dt; much better than Euler's but still the beginning of numerical instability with larger values of Dt

Adding Network Equations Previous slides with the network equations embedded in the differential equations were a special case In general with the explicit approach we'll be alternating between solving the differential equations and solving the algebraic equations Voltages and currents in the network reference frame can be expressed using either polar or rectangular coordinates In rectangular with the book's notation we have

Adding Network Equations Network equations will be written as Y V- I(x,V) = 0 Here Y is as from the power flow, except augmented to include the impact of the generator's internal impedance Constant impedance loads are also embedded in Y; non-constant impedance loads are included in I(x,V) If I is independent of V then this can be solved directly: V = Y-1I(x) In general an iterative solution is required, which we'll cover shortly, but initially we'll go with just the direct solution

Two Bus Example, Except with No Infinite Bus To introduce the inclusion of the network equations, the previous example is extended by replacing the infinite bus at bus 2 with a classical model with Xd2'=0.2, H2=6.0 PowerWorld Case B2_CLS_2Gen

Bus Admittance Matrix The network admittance matrix is This is augmented to represent the Norton admittances associated with the generator models (Xd1'=0.3, Xd2'=0.2) In PowerWorld you can see this matrix by selecting Transient Stability, States/Manual Control, Transient Stability Ybus

Current Vector For the classical model the Norton currents are given by The initial values of the currents come from the power flow solution As the states change (di for the classical model), the Norton current injections also change

B2_CLS_Gen Initial Values The internal voltage for generator 1 is as before We likewise solve for the generator 2 internal voltage The Norton Current Injections are then Keep in mind the Norton current injections are not the current out of the generator

B2_CLS_Gen Initial Values To check the values, solve for the voltages, with the values matching the power flow values

Swing Equations With the network constraints modeled, the swing equations are modified to represent the electrical power in terms of the generator's state and current values Then swing equation is then IDi+jlQi is the current being injected into the network by the generator

Two Bus, Two Generator Differential Equations The differential equations for the two generators are In this example PM1 = 1 and PM2 = -1

Solution at t=0.02 Usually a time step begins by solving the differential equations. However, in the case of an event, such as the solid fault at the terminal of bus 1, the network equations need to be first solved Solid faults can be simulated by adding a large shunt at the fault location Amount is somewhat arbitrary, it just needs to be large enough to drive the faulted bus voltage to zero With Euler's the solution after the first time step is found by first solving the differential equations, then resolving the network equations

Solution at t=0.02 Using Yfault = 1000, the fault-on conditions become

Solution at t=0.02 Then the differential equations are evaluated, using the new voltages and currents These impact the calculation of PEi with PE1=0, PE2=0 If solving with Euler's this is the final state value; using these state values the network equations are resolved, with the solution the same here since the d's didn't vary

Solution Values Using Euler's The below table gives the results using Dt = 0.02 for the beginning time steps Time (Sec) Gen 1 Rotor Angle Gen 1 Speed (Hz) Gen 2 Rotor Angle Gen2 Speed (Hz) 23.9462 60 -12.0829 0.02 60.2 59.9 0.04 25.3862 60.4 -12.8029 59.8 0.06 28.2662 60.6 -14.2429 59.7 0.08 32.5862 60.8 -16.4029 59.6 0.1 38.3462 61 -19.2829 59.5 0.12 45.5462 60.9128 -22.8829 59.5436 0.14 52.1185 60.7966 -26.169 59.6017 0.16 57.8541 60.6637 -29.0368 59.6682 0.18 62.6325 60.5241 -31.426 59.7379 0.2 66.4064 60.385 -33.3129 59.8075 0.22 69.1782 60.2498 -34.6988 59.8751 0.24 70.9771 60.1197 -35.5982 59.9401 0.26 71.8392 59.9938 -36.0292 60.0031 0.28 71.7949 59.8702 -36.0071 60.0649

Solution at t=0.02 with RK2 With RK2 the first part of the time step is the same as Euler's, that is solving the network equations with Then calculate k2 and get a final value for x(t+Dt) Finally solve the network equations using the final value for x(t+Dt)

Solution at t=0.02 with RK2 From the first half of the time step Then

Solution at t=0.02 with RK2 The final value for x(0.02) is Since the d's have changed, the internal voltages are updated, resulting in new Norton currents, and hence new voltages d1(0.02) = 24.7 , d2(0.02) =-12.4

Solution at t=0.02 with RK2 The new values for the Norton currents are

Solution Values Using RK2 The below table gives the results using Dt = 0.02 for the beginning time steps Time (Sec) Gen 1 Rotor Angle Gen 1 Speed (Hz) Gen 2 Rotor Angle Gen2 Speed (Hz) 23.9462 60 -12.0829 0.02 24.6662 60.2 -12.4429 59.9 0.04 26.8262 60.4 -13.5229 59.8 0.06 30.4262 60.6 -15.3175 59.7008 0.08 35.4662 60.8 -17.8321 59.6008 0.1 41.9462 61 -21.0667 59.5008 0.12 48.7754 60.8852 -24.4759 59.5581 0.14 54.697 60.7538 -27.4312 59.6239 0.16 59.6315 60.6153 -29.8931 59.6931 0.18 63.558 60.4763 -31.8509 59.7626 0.2 66.4888 60.3399 -33.3109 59.8308 0.22 68.4501 60.2071 -34.286 59.8972 0.24 69.4669 60.077 -34.789 59.9623 0.26 69.5548 59.9481 -34.8275 60.0267 0.28 68.7151 59.8183 -34.4022 60.0916

Angle Reference The initial angles are given by the angles from the power flow, which are based on the slack bus's angle As presented the transient stability angles are with respect to a synchronous reference frame Sometimes this is fine, such as for either shorter studies, or ones in which there is little speed variation Oftentimes this is not best since the when the frequencies are not nominal, the angles shift from the reference frame Other reference frames can be used, such as with respect to a particular generator's value, which mimics the power flow approach; the selected reference has no impact on the solution